Estimating variance of a Poisson variable The sample variance $S^2$ is an unbiased estimator of the variance $\sigma^2$ of a random variable $X$, and is generally used for this purpose, I believe. But if we assume that $X$ has a Poisson distribution, it seems natural to use the sample mean $\bar{X}$, as $\sigma^2=\lambda=E(X)$. What is the best estimator in this case, and why?
 A: Since Poisson is a member of the regular exponential family, it follows that $\bar X$ is a complete sufficient statistic for $\lambda$. Since $\bar X$ is also unbiased, it follows by the Lehmann–Scheffé theorem that $\bar X$ is the unique minimum variance unbiased estimator (MVUE) of $\lambda$.
Although $S^2$ is unbiased estimator of $\lambda$. Its variance is:
$\mathrm{Var}(S^2) = \frac λn + \frac {2λ^2}{n−1} > \frac λn = \mathrm{Var}(\bar X)$
In fact, after we found MVUE, we do not need to find other estimators, because it is impossible to find the better one.
A: Here's my non mathematical "proof" why the mean estimator must be used.
By definition, you need to know the mean before the variance. That's why the degrees of freedom of the sample variance estimator's distribution is one less than the number of observations: the mean takes away one degree of freedom. Hence, it is impossible to come up with a better estimator of the Poisson's variance than the estimator of its mean, which also happens to be its intensity $\lambda$. So, yes, use the estimator $\hat x=\hat\lambda$ of its mean.
Also, wouldn't it be weird to posit that the variance estimator of Poisson distribution is different than its mean estimator? 
Remark: I wouldn't insist on using $\hat\lambda=\bar x$. It's the best from maximum likelihood point of view, and if that's what you want then great, use it. However, my "proof" is not based on your choice of estimator, particularly $\bar x$. My argument works for any estimator: you can't have a better variance estimator than the estimator of its mean regardless of the choice of the $\hat\lambda$
